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Let $j \in Z_+$. Set $$ a_j^{(1)}=a_j:=\sum_{i=0}^j\frac{(-1)^{j-i}}{i!6^i(2(j-i)+1)!} $$ and $a_j^{(l+1)}=\sum_{i=0}^ja_ia_{j-i}^{(l)}$.

Let $X(i)=|a^{(2i)}_j|j!$. Verify that $X(i)\leq X(1)$ for $j\geq 4i$.

Thank you.

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This seems quite tortuous. Have you made any work or progress yourself? –  Pedro Tamaroff Mar 4 '13 at 4:20
    
    
Nope, in fact, it looks like $X(i) \ge X(1)$ in general. $$\begin{align} X(1) &= ( 1,0,\frac{1}{45},\frac{4}{945},\frac{1}{4725},\frac{8}{18711},\frac{4549}{255405‌​15},\frac{916}{18243225},\frac{28297}{2791213425}, \ldots )\\ X(2) &= ( 1,0,\frac{2}{45},\frac{8}{945},\frac{16}{4725},\frac{256}{93555},\frac{73232}{12‌​7702575},\frac{6592}{18243225},\frac{508672}{1268733375}, \ldots )\\ X(3) &= ( 1,0,\frac{1}{15},\frac{4}{315},\frac{1}{105},\frac{8}{1155},\frac{563}{2837835},‌​\frac{1724}{675675},\frac{61007}{34459425}, \ldots ) \end{align}$$ –  achille hui Mar 4 '13 at 5:06

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